Question

Upon mixing $50.0$ mL of $0.1M$ lead nitrate solution with $50$ mL of $0.05M$ chromic sulphate solution, precipitation of lead sulphate solution takes place. How many moles of lead sulphate are formed and what is the molar concentration of chromic sulphate left in the solution?

• A. $0.005,0.0084$
• B. $0.0084,0.005$
• C. $0.005,0.00084$
• D. $0.05,0.00084$

Answer 1

The correct option is A $0.005,0.0084$

The balanced reaction is as follows:

$3Pb(N{O}_3{)}_2+C{r}_2(S{O}_4{)}_3\to 3PbS{O}_4\downarrow +2Cr(N{O}_3{)}_3$
$3$ mol $Pb(N{O}_3{)}_2\equiv 1$ mol $C{r}_2(S{O}_4{)}_3\equiv 3$ mol $PbS{O}_4$
$50.0$ mL of $0.1M$ lead nitrate solution corresponds to $50.0\times 0.1=5$ mmol.
$50$ mL of $0.05M$ chromic sulphate solution corresponds to $50.0\times 0.05=2.5$ mmol.
$2.5$ mmol of $C{r}_2(S{O}_4{)}_3$ will react with $7.5$ mmol of $Pb(N{O}_3{)}_2$.

However, only $5$ mmol of $Pb(N{O}_3{)}_2$ are present.
Hence, $Pb(N{O}_3{)}_2$ is the limiting reagent.
$5$ mmol of $Pb(N{O}_3{)}_2$ will react with $1.67$ mmol of $C{r}_2(S{O}_4{)}_3$ to give $5$ mmol of $PbS{O}_4$ which corresponds to $0.005$ moles.
Hence, $0.005$ moles of $PbS{O}_4$ are formed.
The mmol of $C{r}_2(S{O}_4{)}_3$ remaining unreacted are $2.5$ mmol $-1.67$ mmol $=0.83$ mmol.
The total volume of the solution is $100$ mL.
The molar concentration of $C{r}_2(S{O}_4{)}_3$ is $\frac{1000}{100}\times \frac{0.83}{1000}=0.0084M$